hdu 5880 Family View ac自动机屏蔽文章关键词 2016青岛网赛

Family View
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 906 Accepted Submission(s): 171

Problem Description
Steam is a digital distribution platform developed by Valve Corporation offering digital rights management (DRM), multiplayer gaming and social networking services. A family view can help you to prevent your children access to some content which are not suitable for them.

Take an MMORPG game as an example, given a sentence T, and a list of forbidden words {P}, your job is to use ‘*’ to subsititute all the characters, which is a part of the substring matched with at least one forbidden word in the list (case-insensitive).

For example, T is: “I love Beijing’s Tiananmen, the sun rises over Tiananmen. Our great leader Chairman Mao, he leades us marching on.”

And {P} is: {“tiananmen”, “eat”}

The result should be: “I love Beijing’s *, the sun rises over *. Our gr*** leader Chairman Mao, he leades us marching on.”

Input
The first line contains the number of test cases. For each test case:
The first line contains an integer
n
, represneting the size of the forbidden words list
P
. Each line of the next
n
lines contains a forbidden words
P
i
(1≤|
P
i
|≤1000000,∑|
P
i
|≤1000000)
where
P
i
only contains lowercase letters.

The last line contains a string
T (|T|≤1000000)
.

Output
For each case output the sentence in a line.

Sample Input
1
3
trump
ri
o
Donald John Trump (born June 14, 1946) is an American businessman, television personality, author, politician, and the Republican Party nominee for President of the United States in the 2016 election. He is chairman of The Trump Organization, which is the principal holding company for his real estate ventures and other business interests.

Sample Output
D*nald J*hn (b*rn June 14, 1946) is an Ame**can businessman, televisi*n pers*nality, auth*r, p*litician, and the Republican Party n*minee f*r President *f the United States in the 2016 electi*n. He is chairman *f The *rganizati*n, which is the p**ncipal h*lding c*mpany f*r his real estate ventures and *ther business interests.

Source
2016 ACM/ICPC Asia Regional Qingdao Online

Recommend
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题意：如题解标题所示。
思路：子串建立ac自动机，并标记各个子串的长度数组，母串在遍历的时候根据在ac自动机查询时候建立的头尾标记选择屏蔽字符。这题需要对trie树和ac自动机扫母串的过程比较了解，如果不懂可以看下这篇文章。[ac自动机原理](http://blog.youkuaiyun.com/niushuai666/article/details/7002823)

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
#define N 100010
#define maxn 1000010
int lay[maxn];//标记母串中子串的出现头和尾
int Len[maxn];
int cnt;
struct Tire
{
//next指某结点相连的26个结点
//fail是tair树上的跳转指针
//End指某结点是否是某个匹配串的结尾
    int next[maxn][26], fail[maxn], End[maxn];
    int root, L;
    int newnode()
    {
        for(int i = 0; i < 26; i++)
        {
            next[L][i] = -1;
        }
        End[L++] = 0;
        return L - 1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }

    void insert(char s[])
    {
        int len = strlen(s);
        int now = root;
        for(int i = 0; i < len; i++)
        {
            if(next[now][s[i] - 'a'] == -1)
            {
                next[now][s[i] - 'a'] = newnode();
            }
            now = next[now][s[i] - 'a'];
        }
        if(End[now]==0)
        {
            End[now] =++cnt;
            Len[cnt]=len;
        }
    }
    void build()  //建立fail指针
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0; i < 26; i++)
            if(next[root][i] == -1)
            {
                next[root][i] = root;
            }
            else
            {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            for(int i = 0; i < 26; i++)
                if(next[now][i] == -1)
                {
                    next[now][i] = next[fail[now]][i];
                }
                else
                {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
        }
    }
    int query(char buf[])
    {
        int len = strlen(buf);
        int now = root;
        int res = 0;
        for(int i = 0; i < len; i++)
        {
            if(buf[i]>='a'&&buf[i]<='z')
                now=next[now][buf[i]-'a'];
            else if(buf[i]>='A'&&buf[i]<='Z')
                now=next[now][buf[i]-'A'];
            else
                continue;
            //now = next[now][buf[i] - 'a'];
            int temp = now;
            while(temp != root)
            {
                lay[i+1]--;
                lay[i-Len[End[temp]]+1]++;
                res += End[temp];
                //End[temp]=0;//只计算在母串中是否出现加上此句
                temp = fail[temp];
            }
        }
        return res;
    }
} ac;
char ps[maxn];
char c[maxn];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        memset(Len,0,sizeof(Len));
        memset(lay,0,sizeof(lay));
        cnt=0;
        ac.init();//初始化
        for(int i=1; i<=n; i++)
        {
            scanf("%s",c);
            ac.insert(c);
        }
        ac.build();//建图
        getchar();
        gets(ps);
        ac.query(ps);
        int sum=0;
        int len=strlen(ps);
        for(int i=0; i<len; i++)
        {
            sum+=lay[i];
            if(sum<=0)
                printf("%c",ps[i]);
            else
                printf("*");
        }
        printf("\n");
    }
}